Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $W$ be a subspace of $V$. Let $W^0 \subset V^*$ be the annihilator of $W$. Prove that $W$ is $T$-invariant if and only if $W^0$ is $T^t$-invariant.
A hint is appreciated for the converse direction.
$W$ is $T$-invariant, then $W^0$ is $T^t$-invariant.
Let $f\in W^0$ (i.e. $f:V\rightarrow K$ such that $f(W)=0$). We want to show $T^t(f)\in W^0$.
Take $w\in W$; by hypothesis $T(w)\in W$ and: $$[T^t(f)](w) = f(T(w)) = 0$$ so $T^t(f)\in W^0$.
$W^0$ is $T^t$-invariant, then $W$ is $T$-invariant.
Let $w\in W$, we want to show $T(w)\in W$.
Take a functional $f\in W^0$; by hypothesis $T^t(f)\in W^0$ and $$f(T(w))=[T^t(f)](w)=0$$ so $T(w)\in \ker f$ for all $f\in W^0$. If we now show that $\bigcap_{f\in W^0}\ker f = W$ we have the statement.
Clearly $W\subseteq \bigcap_{f\in W^0}\ker f$. Vice versa take $x\notin W$, a base $\mathcal B = \{x,v_2,...,v_n\}$ of $V$ and define the functional $f:V\rightarrow K$ such that $$f(x)\neq 0,\quad f(v_2)= \ ... \ =f(v_n)=0$$ $f\in W^0$, and $x\notin \ker f$. Hence $x\notin \bigcap_{f\in W^0}\ker f$ and we are done.